summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of…

Explanation:

Step1: Find the first - derivative

Use the product rule $(uv)' = u'v+uv'$, where $u = 2(4 - x)=8 - 2x$ and $v = e^{x}$. Then $u'=-2$ and $v' = e^{x}$. So $f'(x)=-2e^{x}+(8 - 2x)e^{x}=(6 - 2x)e^{x}$.

Step2: Find the second - derivative

Again use the product rule. Let $u = 6 - 2x$ and $v = e^{x}$, $u'=-2$ and $v' = e^{x}$. Then $f''(x)=-2e^{x}+(6 - 2x)e^{x}=(4 - 2x)e^{x}$.

Step3: Find the inflection points

Set $f''(x)=0$. Since $e^{x}\gt0$ for all $x\in R$, we solve $4 - 2x = 0$, which gives $x = 2$.

Step4: Test the intervals

Choose a test - point in the interval $(-\infty,2)$, say $x = 0$. Then $f''(0)=(4-2\times0)e^{0}=4\gt0$, so the function is concave upward on $(-\infty,2)$. Choose a test - point in the interval $(2,\infty)$, say $x = 3$. Then $f''(3)=(4 - 2\times3)e^{3}=-2e^{3}\lt0$, so the function is concave downward on $(2,\infty)$.

Answer:

B. The function is concave upward on $(-\infty,2)$. It is concave downward on $(2,\infty)$.